---
title: "Does a jump occur at slope change 0.065 to 0.0085"
subject: "Hydraulics"
chapter: 9
key: "semester_4/hydraulics:jump-1-1m-wide-slope-change-location"
source: https://www.bctengineer.com/q/semester_4/hydraulics/jump-1-1m-wide-slope-change-location
site: BCT Engineering
---

# Does a jump occur at slope change 0.065 to 0.0085

Hydraulics · Chapter 9: Non-uniform rapidly varied flow (RVF)

- Asked in: 2071 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/jump-1-1m-wide-slope-change-location
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-9

## Question

A rectangular channel with width 1.1 m carrying a flow discharge of 7.2 m³/s changes its bed slope from 0.065 to 0.0085. Show that the hydraulic jump occurs and if so find the location of jump. Take Manning's roughness as 0.025.

## Short answer

Normal depths: 1.224 m supercritical and 2.962 m subcritical. Sequent of 1.224 m is 2.128 m, less than downstream depth, so jump moves upstream. S1 backwater from break to 2.128 m gives jump location ≈ 11.4 m upstream of slope change.

## Answer

**Given:** $b = 1.1$ m, $Q = 7.2\ \text{m}^3/\text{s}$ ($q = 6.545\ \text{m}^2/\text{s}$), $n = 0.025$, $S_{01} = 0.065$ changing to $S_{02} = 0.0085$.

$$
y_c = \left(\frac{6.545^2}{9.81}\right)^{1/3} = 1.635\ \text{m}
$$

**Normal depths** (Manning's equation with $R = A/P$, by trial):

| Reach | $S_0$ | $y_n$ (m) | $V$ (m/s) | $Fr$ | Flow |
|---|---|---|---|---|---|
| 1 | 0.0650 | 1.224 | 5.35 | 1.54 | supercritical (steep) |
| 2 | 0.0085 | 2.962 | 2.21 | 0.41 | subcritical (mild) |

Flow on the upstream reach is supercritical ($y_{n1} < y_c$) and on the downstream reach it is subcritical ($y_{n2} > y_c$). The change from supercritical to subcritical flow can occur only through a **hydraulic jump**, so a jump occurs.

#### Location

Sequent depth of $y_{n1} = 1.224$ m:

$$
y_2 = \frac{1.224}{2}\left(\sqrt{1 + 8(1.54)^2} - 1\right) = 2.128\ \text{m}
$$

Because $y_2 = 2.128\ \text{m} < y_{n2} = 2.962$ m, the downstream depth is higher than needed. The jump is pushed **upstream of the break**, onto the steep slope. After the jump the depth rises from 2.128 m to 2.962 m along an **S1** curve ending at the break. The jump forms where the S1 depth equals 2.128 m.

**Direct step on the S1 curve** (from 2.962 m at the break to 2.128 m, 4 steps), with $S_0 = 0.065$:

| $y$ (m) | $R$ (m) | $V$ (m/s) | $E$ (m) | $S_f$ | $\bar S_f$ | $\Delta x$ (m) |
|---|---|---|---|---|---|---|
| 2.962 | 0.4639 | 2.210 | 3.2108 | 0.00850 | | |
| 2.753 | 0.4584 | 2.377 | 3.0415 | 0.00999 | 0.00925 | -3.04 |
| 2.545 | 0.4523 | 2.572 | 2.8821 | 0.01191 | 0.01095 | -2.95 |
| 2.336 | 0.4452 | 2.801 | 2.7365 | 0.01443 | 0.01317 | -2.81 |
| 2.128 | 0.4370 | 3.076 | 2.6102 | 0.01783 | 0.01613 | -2.58 |

$$
L = 3.04 + 2.95 + 2.81 + 2.58 \approx 11.4\ \text{m}
$$

**Answer: A hydraulic jump occurs on the steep reach about 11.4 m upstream of the slope change (depths 1.224 m to 2.128 m), followed by an S1 curve up to the break.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why does a jump occur?
   A. Upstream normal depth is above critical depth
   B. Both reaches are mild
   C. Upstream flow is supercritical (yn1 < yc) and downstream is subcritical (yn2 > yc)
   D. The discharge changes
   Answer: C. Supercritical to subcritical flow can only change through a jump. (Key point: Does a jump occur)

2. Why does the jump occur on the steep reach rather than downstream of the break?
   A. Sequent depth 2.128 m is less than yn2 = 2.962 m, so the tailwater is more than enough and pushes the jump upstream
   B. The sequent depth is more than yn2
   C. The slope is mild
   D. Because friction is zero
   Answer: A. High tailwater drives the jump upstream, followed by an S1 curve up to the break. (Key point: Location)

3. How is the jump position found?
   A. Direct step along the S1 curve from 2.962 m to 2.128 m, giving about 11.4 m
   B. Using the jump length 6(y2 - y1)
   C. By Belanger equation only
   D. By the critical slope
   Answer: A. The jump forms where the S1 depth equals the sequent depth 2.128 m. (Key point: Distance)

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